V1 = 2.00 L
<span>T1 = 25 + 273 = 298 K </span>
<span>V2 = 6.00 L </span>
<span>T2 = ? </span>
<span>Assuming the pressure is to remain constant, then </span>
<span>V1/T1 = V2/T2 </span>
<span>T2 = T1V2/V1 = (298)(6)/(2) = 894 deg K</span>
We have been given the condition that carbon makes up 35%
of the mass of the substance and the rest is made up of oxygen. With this, it
can be concluded that 65% of the substance is made up of oxygen. If we let x be
the mass of oxygen in the substance, the operation that would best represent
the scenario is,
<span> x = (0.65)(5.5 g)</span>
<span> <em> </em><span><em>x =
3.575 g</em></span></span>
"CH3COOH + H2O CH3COO- + H3O+" is the equation among the choices given in the question <span>represents the reaction of acetic (ethanoic) acid with water. The correct option among all the options that are given in the question is the second option or option "B". I hope that the answer has helped you.</span>
The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
